Integral of $(z^2 + x^2)^{-\frac{3}{2}}$

Question

I am studying Griffiths Introduction to Electrodynamics, in which the following integral appears:

$$\int_{-L}^L\left({z^2+x^2}\right)^{-\frac{3}{2}}\,dx$$

where $z$ denotes a constant, and $z\in \mathbb{R}$

The integration is done without much fuzz about it, as if it was simple. I tried throwing substitution, and partial integration at it, which didn't bring me far.

I now stumbled upon the following video https://www.youtube.com/watch?v=KR22vdvL_3g which deals with a similar integral. There, use is made of hyperbolic functions and some rather advanced looking identities. The speaker also mentions that this integral was particularly nasty to solve.

I find it hard to believe that one of the first problems in a rather introductory text should be so difficult, especially because the integration is done in one step, without any explanation.

Are there simpler methods of doing this integral, ones that i am missing? If not, i guess my question becomes: How to solve this integral?

Answer 1

Substitute $x=z\tan u\implies dx=z\sec^2udu$: $$I=\int_{-\arctan \frac Lz}^{\arctan \frac Lz} \frac{z\sec^2u \ du}{z^3\sec^3 u} =\frac{2}{z^2}\int_0^{\arctan \frac Lz}\cos u du\\ =\frac{2}{z^2} \sin\left(\arctan \frac Lz\right)\\=\frac{2}{z^2} \frac{\frac Lz}{\sqrt{1+\frac{L^2}{z^2}}}\\=\frac{2L}{z^2\sqrt{L^2+z^2}}$$

Answer 2

Assuming that $z$ is a (real) constant. Observe that the function $f(x)=\frac{1}{\sqrt{(x^2+z^2)^3}}$ is an even function. So \begin{align*} I & = \int_{-L}^L \frac{1}{\sqrt{(x^2+z^2)^3}} \, dx\\ &= 2\int_{0}^L \frac{1}{\sqrt{(x^2+z^2)^3}} \, dx\\ &= \frac{2}{z^2}\int_{0}^{\arctan\frac{L}{z}} \cos u\, du && (\text{ let } x =z \tan u ) \end{align*} Now this should be easy to solve.